99

I have a string which holds a decimal value in it and I need to convert that string into a floating point variable. So an example of the string I have is "5.45" and I want a floating point equivalent so I can add .1 to it. I have searched around the internet, but I only see how to convert a string to an integer.

Vadim Kotov
  • 8,084
  • 8
  • 48
  • 62
Anton
  • 12,285
  • 20
  • 64
  • 84
  • The question needs further detail. Is the string only composed of numerals? Alphanumeric? Ok if alphas are stripped? Is there a specific purpose for the resulting number? – JGurtz Mar 01 '16 at 22:51

9 Answers9

96

You don't need to convert it at all:

% perl -e 'print "5.45" + 0.1;'
5.55
Alnitak
  • 334,560
  • 70
  • 407
  • 495
  • 15
    @OrangeDog the OP edited the question (some months after this answer was posted) - the original question actually had floating point numbers in it. – Alnitak Mar 14 '12 at 10:46
  • 1
    what about comparisons when the string has a comma in it? – Ramy May 22 '14 at 15:21
  • `[rabdelaz@Linux_Desktop:~/workspace/akatest_5]$perl -e 'print "nope\n" unless "1,000" > 10;' nope [rabdelaz@Linux_Desktop:~/workspace/akatest_5]$perl -e 'print "nope\n" if "1,000" > 10;'` – Ramy May 22 '14 at 15:24
79

This is a simple solution:

Example 1

my $var1 = "123abc";
print $var1 + 0;

Result

123

Example 2

my $var2 = "abc123";
print $var2 + 0;

Result

0
Anne
  • 26,765
  • 9
  • 65
  • 71
porquero
  • 1,159
  • 8
  • 5
44

Perl is a context-based language. It doesn't do its work according to the data you give it. Instead, it figures out how to treat the data based on the operators you use and the context in which you use them. If you do numbers sorts of things, you get numbers:

# numeric addition with strings:
my $sum = '5.45' + '0.01'; # 5.46

If you do strings sorts of things, you get strings:

# string replication with numbers:
my $string = ( 45/2 ) x 4; # "22.522.522.522.5"

Perl mostly figures out what to do and it's mostly right. Another way of saying the same thing is that Perl cares more about the verbs than it does the nouns.

Are you trying to do something and it isn't working?

Rop
  • 3,359
  • 3
  • 38
  • 59
brian d foy
  • 129,424
  • 31
  • 207
  • 592
  • 1
    Forgive my lack of knowledge here but I don't quite get your second example. You're dividing two numbers, then multiplying them, how/why does this result in a string? – gideon Nov 28 '12 at 14:35
  • 4
    I'm not multiplying numbers. The `x` is the string replication operator. – brian d foy Jan 07 '13 at 09:12
  • 3
    Shouldn't it be `my $string = ( 45/2 ) x 3; # "22.522.522.5"` with 45 instead of 44? Otherwise I don't get where the '.5's come from in the result... – Vickster Feb 27 '13 at 13:42
10

Google lead me here while searching on the same question phill asked (sorting floats) so I figured it would be worth posting the answer despite the thread being kind of old. I'm new to perl and am still getting my head wrapped around it but brian d foy's statement "Perl cares more about the verbs than it does the nouns." above really hits the nail on the head. You don't need to convert the strings to floats before applying the sort. You need to tell the sort to sort the values as numbers and not strings. i.e.

my @foo = ('1.2', '3.4', '2.1', '4.6');
my @foo_sort = sort {$a <=> $b} @foo;

See http://perldoc.perl.org/functions/sort.html for more details on sort

Norm
  • 113
  • 2
  • 8
9

As I understand it int() is not intended as a 'cast' function for designating data type it's simply being (ab)used here to define the context as an arithmetic one. I've (ab)used (0+$val) in the past to ensure that $val is treated as a number.

brian d foy
  • 129,424
  • 31
  • 207
  • 592
mccutchm
  • 139
  • 1
  • 3
7
$var += 0

probably what you want. Be warned however, if $var is string could not be converted to numeric, you'll get the error, and $var will be reset to 0:

my $var = 'abc123';
print "var = $var\n";
$var += 0;
print "var = $var\n";

logs

var = abc123
Argument "abc123" isn't numeric in addition (+) at test.pl line 7.
var = 0
indexless
  • 389
  • 3
  • 11
5

Perl really only has three types: scalars, arrays, and hashes. And even that distinction is arguable. ;) The way each variable is treated depends on what you do with it:

% perl -e "print 5.4 . 3.4;"
5.43.4


% perl -e "print '5.4' + '3.4';"
8.8
brian d foy
  • 129,424
  • 31
  • 207
  • 592
Rini
  • 197
  • 2
  • 7
1

In comparisons it makes a difference if a scalar is a number of a string. And it is not always decidable. I can report a case where perl retrieved a float in "scientific" notation and used that same a few lines below in a comparison:

use strict;
....
next unless $line =~ /and your result is:\s*(.*)/;
my $val = $1;
if ($val < 0.001) {
   print "this is small\n";
}

And here $val was not interpreted as numeric for e.g. "2e-77" retrieved from $line. Adding 0 (or 0.0 for good ole C programmers) helped.

memowe
  • 2,656
  • 16
  • 25
Steffen Moeller
  • 11
  • 1
1

Perl is weakly typed and context based. Many scalars can be treated both as strings and numbers, depending on the operators you use. $a = 7*6; $b = 7x6; print "$a $b\n";
You get 42 777777.

There is a subtle difference, however. When you read numeric data from a text file into a data structure, and then view it with Data::Dumper, you'll notice that your numbers are quoted. Perl treats them internally as strings.
Read:$my_hash{$1} = $2 if /(.+)=(.+)\n/;.
Dump:'foo' => '42'

If you want unquoted numbers in the dump:
Read:$my_hash{$1} = $2+0 if /(.+)=(.+)\n/;.
Dump:'foo' => 42

After $2+0 Perl notices that you've treated $2 as a number, because you used a numeric operator.

I noticed this whilst trying to compare two hashes with Data::Dumper.

SzG
  • 12,333
  • 4
  • 28
  • 41